Mathematics - 2018-02-23 (page 3 of 4)

Akiva Weinberger

12:17

If $f(x)$ is convex, $f(x)+x$ is convex, right? Not assuming differentiability

Right, if and only if, even

Balarka Sen

sum of two convex functions is convex my man

Akiva Weinberger

Right yeah

and $x$ and $-x$ are convex

I forgot

Secret

y = 0 is convex?

Akiva Weinberger

But not the product of two convex functions

@Secret Not strictly, but yeah

Balarka Sen

If $\gamma$ is a closed arclength parametrized planar curve then the total curvature $\int_\gamma K d\ell = 2\pi$, right? I forget what's the proof of this.

Maybe it's Gauss-Bonnet theorem on the domain the $\gamma$ bounds.

Akiva Weinberger

12:27

If it intersects itself it could be $4\pi$

Balarka Sen

That must be it. $\chi(D^2) = 1$

Akiva Weinberger

Of zero, I think a figure-8 should have 0

Balarka Sen

@Akiva Yeah I meant embedded

a Jordan curve

Akiva Weinberger

Right then yeah it should be

A smooth Jordan curve, so you can measure the curvature

Balarka Sen

Mhm

Or at least $C^2$

Akiva Weinberger

12:28

Right

Well, $\pm2\pi$ :P

Depending on how you orient it

Balarka Sen

Bleh :P

You're right, I agree

Akiva Weinberger

I don't know the names of the theorems but I'm sure you could regular-homotopy it to a circle

Balarka Sen

You can but proving that preserves total curvature seems like a pain. It's just easier to just apply G-B

Akiva Weinberger

But that just tells you it's $2\pi k$ for some $k$, doesn't it?

Balarka Sen

Uh, no, I get $2\pi \chi(D^2)$

because by Jordan Schoenflies it bounds a disk

and euler char of disk is 1

Akiva Weinberger

12:33

A diffeomorphic disk, even, I think

and this should apply in all dimensions 'cause smoothness stuffs

Balarka Sen

The left hand side is integral of Gaussian curvature of disk + integral of geodesic curvature of $\gamma$

the first integral vanishes

and geodesic curvature here is exactly $K$

Silent

Please suggest me a link or literature for this comment. Also, what does bounded mean there?

Akiva Weinberger

There's actually two generalizations of this to higher dimensions, I realize

One is that an embedded sphere has total curvature $4\pi$, I think it is

Balarka Sen

Ok, there's a theorem which says if $\gamma$ is an arclength parametrized nontrivial knot in $\Bbb R^3$ then $\int_\gamma K d\ell > 4\pi$. In light of this proof, I wonder if doing something similar with a Seifert surface bounding $\gamma$ works

Akiva Weinberger

and the second is that an embedded circle in $\Bbb R^3$ has total curvature $>2\pi$, and if it's knotted it has total curvature $>4\pi$ (but it's no longer quantized)

Balarka Sen

12:36

@AkivaWeinberger That's just Gauss-Bonnet on the sphere

Akiva Weinberger

Right

Balarka Sen

w/ the metric coming from pulling back the metric on R^3 by the embedding

Akiva Weinberger

Oh wait

Hold on, that is true for all immersions of the sphere, isn't it

Balarka Sen

Yes

Akiva Weinberger

despite the equivalent statement with circles in the plane not being true

because you can get all sorts of [forgetting the name of it] winding numbers of the tangent vector

…Turning number, that's it

Balarka Sen

12:38

Well that's because the curvature of a curve is not analogous to curvature of an embedded sphere

Akiva Weinberger

No?

Balarka Sen

curvature of a curve is obtained very extrinsically, whereas the curvature of an immersed sphere is obtained by pulling back the metric to S^2 and then computing the curvature of the metric there

I mean it's true that for any metric on S^2, curvature has to be $4\pi$

Akiva Weinberger

But you can also compute the curvature of an immersed sphere extrinsically

Balarka Sen

Regardless of how the metric is obtained

Akiva Weinberger

can't you?

Balarka Sen

12:40

When you immerse a sphere the curvature is an intrinsic property of the pullback metric

Akiva Weinberger

@Silent The reasoning is explained in the edited version of the answer given

"Bounded" meaning you can't do, like, $(1-1)+(\frac12+\frac12-\frac12-\frac12)+(\frac13+ \frac13+\frac13- \frac13-\frac13-\frac13)+\dotsb$

because that series (without the parentheses) clearly diverges (as its partial sums keep hitting 1), despite the fact that the series with parentheses is 0 and that the terms go to zero

Balarka Sen

@AkivaWeinberger You can compute it extrinsically, but it is not an extrinsic property.

That's what Gauss's theorema egregium is about

Akiva Weinberger

If you have a limit on the number of terms that can be grouped, that can't happen

@BalarkaSen I see

Weird that it's so different from the 1D case

Balarka Sen

Yeah

Akiva Weinberger

This continues for higher dimensions as well, right?

That it's intrinsic?

Balarka Sen

12:44

You have to be careful about what curvature is then

The right generalization is the Riemann curvature tensor

Which is not a number anymore

but a tensor of numbers

It is intrinsic, yes. I think there's ways to compute it extrinsically too, but I do not know the theory. Ted definitely can tell you about this

Akiva Weinberger

Oh, what?

Balarka Sen

@AkivaWeinberger I can explain what it is in a few lines if you want

Akiva Weinberger

Why can't we define it by a limit of angle excess/defect?

Oh, wait, but angle isn't really defined in higher dimensions

Balarka Sen

Yeah, so you have to be careful

Akiva Weinberger

There's no generalization of the theorem that the sum of the angles of a plane triangle are 180, in higher dimensions, is there

Balarka Sen

12:47

Not that I know of, but maybe there is

Secret

not even in terms of solid angles?

I would imagine a tetrahedron sum of solid angles should be something well defined?

4

Q: Sum of dihedral angles in Tetrahedron

I'd like to ask if someone can help me out with this problem. I have to determine what is the lower and upper bound for sum (the largest and smallest sum I can get) of dihedral angles in arbitrary Tetrahedron and prove that. I'm ok with hint for proof, but I'd be grateful for lower and upper boun...

geometry

ah, it's $2\pi$

Balarka Sen

@AkivaWeinberger Let me tell you about the exponential map first

$(M, g)$ be a Riemannian manifold and $p \in M$ be a point

You can look at the various geodesics passing through $p$

Akiva Weinberger

Maybe we need some version of, like, the sum of the angles times the side lengths

Balarka Sen

@Akiva I think you're going to end up defining the Ricci curvature but don't quote me on that

Akiva Weinberger

So is this gonna be like the azimuthal projection

Balarka Sen

12:51

Anyhow, recall that geodesics passing through $p$ with velocity vector $v \in T_p M$ are solutions to $\nabla_{\gamma'} \gamma' = 0$ with initial condition $\gamma(0) = p$ and $\gamma'(0) = v$ where $\nabla$ is the unique metric connection satisfying symmetry like we discussed earlier

Akiva Weinberger

AKA map of the Flat Earth wake up sheople

Balarka Sen

@Akiva I'm actually going to describe that picture to you

Akiva Weinberger

@BalarkaSen Right

Balarka Sen

So that's an ODE with fixed initial conditions.

Picard-Lindelof guarantees that there is a unique solution

That means each vector $v \in T_p M$ determines a unique geodesic through $p$ with velocity vector $v$

(Think, eg, of the sphere. Every direction at the north pole determines a unique geodesic/great circle through it. If you take a longer direction, the speed of the geodesic multiplies, so it's still a different geodesic)

Akiva Weinberger

Woo I don't have class first period

I'm pretty sure that means I can come 40 minutes late

Balarka Sen

12:55

\o/

Akiva Weinberger

That's good because I am somewhere around 40 minutes late already by accident

@BalarkaSen I can believe this

So then you map a $(r,\theta)$ in the tangent plane to the thingy?

Balarka Sen

@Akiva Yup. That means you have a map $\text{exp} : T_p M \to M$ sending each vector $v \in T_p M$ to $\gamma(1)$ where $\gamma$ is the unique geodesic through $p$ with velocity vector $v$.

This is the so-called exponential map

Except I'm lying a little and $\exp$ is only defined on a small neighborhood of the zero vector on $T_p M$.

Otherwise solutions to the ODE may "blow up" for large velocities

Secret

@BalarkaSen actually, my link might be irrelevant because you might be looking for a result of plane angles generalised to higher dimensional "corners"

Balarka Sen

In any case $\exp$ is still defined everywhere if $M$ is a compact manifold so you can forget about that

$\exp$ is also a diffeomorphism near a neighborhood of zero

So it's like mapping a tiny patch of the tangent space to a tiny patch near $p$ in $M$

Akiva Weinberger

@BalarkaSen Is this 'cause you can fall into a whole again?

Like think of the exponential map of a plane minus a closed disk somewhere

Balarka Sen

13:01

Yep, that's it!

This happens when the manifold is "not complete"

Akiva Weinberger

Mhm

With "complete" in the metric spaces sense of "complete"?

Balarka Sen

Right :)

I can tell you more about $\exp$ later but here's a way to define the Riemann curvature tensor from this

Suppose $B \subset T_p M$ is a ball around the zero vector such that $\exp_p : B \to \exp_p(B) \subset M$ is a diffeomorphism

Akiva Weinberger

Sure

Balarka Sen

For any 2-plane $S$ in $T_p M$, consider $\exp_p(B \cap S)$. This is a "geodesic surface" in $M$ passing through $p$.

THIS has a well defined Gaussian curvature $K_S$

Akiva Weinberger

Wait, how many dimensions is $M$?

Balarka Sen

13:05

Greater than 2 for now

Akiva Weinberger

Oh, so this 2-plane doesn't need to be a hyperplane (codimension 1).

Balarka Sen

Yep

It's just a dim 2 surface patch inside $M$ at $P$

Akiva Weinberger

Mhm, and now we have geodesic surfaces in M with curvatures.

Balarka Sen

Right

So we have a function $\text{Gr}_2(T_pM) \to \Bbb R$ from the Grassmannian of 2-planes in $T_p M$

that takes a 2-plane, projects it to a geodesic surface in $M$ by exponential map and takes the Gaussian curvature at $p$

Akiva Weinberger

Sure

Balarka Sen

13:08

Choose a basis $e_1, e_2, \cdots, e_n$ of $T_p M$ and define $K(e_i, e_j)$ to be the curvature of $\exp_p(\text{Span}(e_1, e_2) \cap B)$ at $p$

Secret

Generalized Gauss–Bonnet theorem

In mathematics, the generalized Gauss–Bonnet theorem (also called Chern–Gauss–Bonnet theorem after Shiing-Shen Chern, Carl Friedrich Gauss and Pierre Ossian Bonnet) presents the Euler characteristic of a closed even-dimensional Riemannian manifold as an integral of a certain polynomial derived from its curvature. It is a direct generalization of the Gauss–Bonnet theorem (named after Carl Friedrich Gauss and Pierre Ossian Bonnet) to higher dimensions. Let M be a compact orientable 2n-dimensional Riemannian manifold without boundary, and let Ω {\displaystyle...

Balarka Sen

This gives a matrix of Gaussian curvatures - that's precisely the Riemann curvature tensor more or less

Secret

higher dimensions

Balarka Sen

(Actually it's defined as $R_{ijkl} = g_p(K(e_i, e_j)e_k, e_l)$ but whatever)

@Secret I should learn the proof of that theorem sometime soon

Ted sent me some notes on it...

Akiva Weinberger

@BalarkaSen Interesting

Balarka Sen

13:18

What's weird is that Riemann's original definition of the curvature tensor is super strange

It's defined as $R(X, Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X, Y]} Z$

Gromov famously said nobody understands this definition

It's very handy but geometrically not insightful

Akiva Weinberger

Why did he define that then

Balarka Sen

One explanation is that it's an infinitisimal parallel transport around the square "$XYX^{-1}Y^{-1}$"

But I don't buy it

0celo7

gromov is full of shit

that equation is the integrability condition for a space to be locally Euclidean

Secret

I saw the same description in sean caroll's lecture notes on GR when he describe what the curvature tensor means

0celo7

that's what curvature means, anyone saying anything else is high on krokodil

Balarka Sen

13:23

anything can be a integrability condition. what does that have anything to do with being curved?

i understand your explanation but it's not very good

$d\omega = 0$ is an integrability condition for almost symplectic structures to be symplectic. What does that mean?

It doesn't provide any geometric insight

0celo7

the opposite of flat is curved, and if it's not isometric to flat space it should be the opposite of flat

Balarka Sen

Curvature should be way more than just an obstruction to being (locally) isometric to Euclidean space

skullpatrol

"flat" doesn't have a "single" opposite

0celo7

But that's exactly what it is, locally.

skullpatrol

what's the opposite of "parallel"?

philmcole

13:28

If I have a set of vectors $v_1,v_2,v_3,v_4$ and I want to know if they are linearly dependent, I can put them as columns in a matrix $A$ and bringt it into reduced row echelon form so I can read off the coefficients with which they form a linear combination that gives zero, right? So assuming I get such a linear combination and decide to throw away $v_4$ because it is redundant. Do I have to now do everything again to determine if the remaining three are linearly dependent or not?

Is there not a quick way of throwing out all redundant vectors in a set and keeping only the minimal set?

Balarka Sen

@0celo7 idk man I love to think about curvature as a measure of how fast two geodesics converge or diverge away from themselves

maybe I'm i n f l u e n c e d b y G r o m o v

0celo7

that's a perfectly reasonable definition too

I just think it's completely wrong to say no one understands that equation -- I think I understand it.

Balarka Sen

you know Gromov

skullpatrol

he's playing with his personal definition of "understanding"

Secret

Also, just some amendment to what I said earlier: The sum of solid angle result is actually not $2\pi$ but bounded above by $2\pi$, thus it is still different from the 2D case where a flat triangle must have all angles sum to $\pi$

Anyting else atm I don't know enough to comment

Akiva Weinberger

13:42

Can you measure the angles at the vertices?

I mean, sure, you can define something pretty easily probably

Does it have a name?

Secret

Well... if you mean 2D angles, and say the vertex has n faces connected to it, then you have n angles (not including those that go from one edge to another that is not on a face)

Akiva Weinberger

Oh, wait, what I'm thinking of is the same as the angle deficiency I think

Oh wow that's cool actually

I have two completely different definitions in my head and they seem to agree

Right, so, consider a vertex of a cube

and draw a small sphere of radius $r$ around that vertex

Secret

Akiva Weinberger

The interior of that cube will intersect 1/8 of the surface of that sphere, right?

Secret

yes, and that's what we call a solid angle

Akiva Weinberger

13:46

That sphere has $4\pi$ area, $4\pi/8$ is $\pi/2$

So through that definition I'll say that the vertex has angle $\pi/2$ (note that this is defined by the area of a 2D thing)

A second definition: The three faces next to the vertex each have a $\pi/2$ plane angle on it

with a total of $3\pi/2$

Secret

In fact, it has units $\frac{\pi}{2}$ steradians

Akiva Weinberger

Oh wait hold on never mind

I was gonna subtract it from $2\pi$ and get the angle deficiency but this actually is gonna be different from the first thing in general

Secret

I think for the second definition, you will be capturing how much that corner deviate from being planar, which could be one of the components of the curvature tensor

Meanwhile, I don't really know whether there is a clean relation between solid angles and curvature

Akiva Weinberger

It is, it's essentially the curvature of the vertex (divided by Dirac delta)

Secret

Actually, if you do $2\pi - \frac{3\pi}{2}$ you get $\frac{\pi}{2}$. While it has he same value as the solid angle, I will be hesitant without further checking to say they are not just coincidence

Spherical trigonometry

Spherical trigonometry is the branch of spherical geometry that deals with the relationships between trigonometric functions of the sides and angles of the spherical polygons (especially spherical triangles) defined by a number of intersecting great circles on the sphere. Spherical trigonometry is of great importance for calculations in astronomy, geodesy and navigation. The origins of spherical trigonometry in Greek mathematics and the major developments in Islamic mathematics are discussed fully in History of trigonometry and Mathematics in medieval Islam. The subject came to fruition in Early...

Th following may be relevant to the discussion

there's a section that talks about how much an angle exceed $\pi$ radians

and in the solid angle article, it seems the dihedral angle of all the faces around the vertex is enough to define it. But i have to check closer to better understood it

Semiclassical

13:57

I was actually playing around with spherical trig stuff a while back, though it turned out to be overkill

Secret

I don't know much spherical geomtry, other than how dihedrals are measured and that internal angles of spherical triangels sum to > $\pi$. This is because in catalysis, there is a notion of biting angle which basically tells you how big is the volume enclosed by the surrounding ligands, which is given by the largest cone the ligands can substend

philmcole

Does somebody have an idea? chat.stackexchange.com/transcript/message/43025736#43025736

Secret

6

Q: How fast can you determine if vectors are linearly independent?

Let us suppose you have $m$ real-valued vectors of length $n$ where $n \geq m$. How fast can you determine if they are linearly independent? In the case where $m = n$ one way to determine independence would be to compute the determinant of the matrix whose rows are the vectors. I tried some go...

linear-algebra

philmcole: I don't think there is any quick way. Even proofs in linear algebra that require producing a linear independent subset either use gram schmidt or in the proof, has the notion of throwing away dependent vectors one by one

Semiclassical

Suppose you’ve got three unit vectors; there will be three pairwise dot products, which together correspond to a point in [-1,1]^3. So there’s a mapping from such triplets of unit vectors to the cube, and I wanted to know what the range of this map was.

philmcole

@Secret Thanks! I'll keep that in mind. Probably that's why we deal only with a small set of like three vectors in most examples in class because there is no way of quickly determining how many of 17 vectors are redundant and can be thrown away.

Secret

14:07

0

Q: Extracting the largest set of linearly independent vectors from a set of vectors in matlab

I am new to working with matlab. I couldn't find an obvious way to to extract the largest subset of linearly independent vectors from a given set of vectors. So given a set V = [v1 v2 -- vn] where dim(vi) >> n (i=1,2,3,....) I need to find a random set of r linearly independent vectors "vi" (w...

matlab vector

Actually, it seems my linear algebra is getting rusty, and I forgot that the row echelon form has information on what the column space of the matrix is

@Semiclassical I don't understand, do you mean mapping the triplets to the corners of the cube, or also other edges. Also if the triplet get mapped to a vertex, are they all pointing outward or you also want all posible permutation of pointing directions?

Semiclassical

well, not all 6 corners of the cube can be realized

If you pick the corner (1,1,1) that corresponds to the dot products all become 1 ie the unit vectors are all parallel. No problem with that

However, the point (1,1,-1) would mean one vector is parallel to the other two but these two are antiparallel

So that corner isn’t going to be in the range since no triplet of unit vectors can possibly generate it

Equivalently, you could just work with the angles directly. In that case you want to know what sets of relative angles can be formed from unit vectors

But I can associate a unit vector in RR^3 to a point on the unit sphere S^2, and then the angle between two such points corresponds to the distance between them on the unit sphere

Secret

Actually, I felt like I need a lot more information: First, are the vectors all parallel or coplanar to each other, cause from what you said about the corners, it seems my initial presumption that they are orthogonal to each other is wrong?

Semiclassical

So the three angles correspond to the side lengths of the corresponding spherical triangle

@Secret the vectors are a set of three unit vectors in RR^3

And the initial mapping is from these vectors to the three pairs of dot products between them

Akiva Weinberger

49 mins ago, by philmcole

If I have a set of vectors $v_1,v_2,v_3,v_4$ and I want to know if they are linearly dependent, I can put them as columns in a matrix $A$ and bringt it into reduced row echelon form so I can read off the coefficients with which they form a linear combination that gives zero, right? So assuming I get such a linear combination and decide to throw away $v_4$ because it is redundant. Do I have to now do everything again to determine if the remaining three are linearly dependent or not?

But the remaining three are in echelon form now, aren't they?

So going from there shouldn't be too hard

@philmcole

philmcole

True. Btw could you figure out how many vectors to throw away by the number of free variables you get when reducing A? So for example if we get the two free variables $x_3,x_4$ then $\vec{x} = x_3 \vec{c_1} + x_4 \vec{c_1}$ is a solution of $A \vec{x} = \vec{0}$.

Secret

14:22

@Semiclassical right, so the mutual orthogonal case will all map to the origin, hmm...

Semiclassical

yep

What this leads to is: What is the spherical version of the triangle inequality?

To make things simple, I'll assume that the angles i.e. side lengths are all $\leq \pi$

so if I pick a two points on the sphere and draw a great circle through them, the side length is the smaller of the two arcs on the great circle.

(You don't need this if you work with dot products since $\cos(2\pi-\theta)=\cos \theta$. But for angles it's easier if you can restrict yourself to $[0,\pi]$.)

Akiva Weinberger

@Secret Here's something interesting, though I can't remember if I shared it here before

So it's about something that works in 2D but not 3D

Semiclassical

The restrictions turn out to be rather simple. First, you can't have $\theta_{ab}+\theta_{bc}<\theta_{ac}$ where $a,b,c$ are your three unit vectors

Akiva Weinberger

Consider the set of all lattice polygons, and the area function from that set to $\Bbb R_{\ge0}$

Notice that it's invariant under translation and rotation, and that it's additive

Secret

@Semiclassical I am still thinking, but one thing we are at least sure about is that the (a,b,c) cannot be of permutations (1,1,-1), (1,-1,-1). That actually rules out a lot of corners of the cube, leaving behind only $\pm$(1,1,1)

Semiclassical

14:29

@Secret Not quite: You can have (1,-1,-1) and its permutations

Akiva Weinberger

First, prove that anything translation invariant and additive on the set of lattice polygons is necessarily a constant multiple of the area

Semiclassical

That'd mean two vectors are parallel and the last is antiparallel to both of them. But that's easy enough: two vectors pointing up and one pointing down.

So you have half of the six corners being allowed and half being ruled out.

Arnoud Nijon

Quick terminology question: I don't know how I should call the process of obtaining a morphism from another morphism through an adjunction. 'Performing the adjunction on f, we obtain g'. Or, 'we obtain g as the adjunctional transpose of f'. Any ideas?

Semiclassical

To make this more algebraic, let $L=[a\, b\, c]$ where $a,b,c\in\mathbb{R}^3$ are our unit vectors.

Secret

Guys, I am going to take a bath now. I will get back to all of you in a few minutes to hous

Semiclassical

14:33

Then the matrix $L^T L$ is the matrix of pairwise inner products. The matrix is obviously symmetric, and the diagonal elements are ones since we're working with unit vectors.

bai

Secret

keep posting while I am afking

Semiclassical

kk

So the matrix $L^T L$ is determined by its three off diagonal elements. But not all symmetric matrices with ones on the diagonal can arise this way; for instance, you can't obtain $$L^T L=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1\end{pmatrix}$$ in this way

Actually, I'm going to rename $L\to U$. I meant to do that and accidentally stuck with an obsolete convention.

also, the off-diagonal elements of $U^T U$ should be dot products and therefore in [-1,1]

so a generic such matrix being $M=\begin{pmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1\end{pmatrix}$ and $(x,y,z)\in[-1,1]^3$

so one wants to know when $M=U^T U$ with the columns of $U$ being unit vectors.

Secret

just back, now analysing...

Semiclassical

14:48

Easiest way I know to proceed is to note that if $M=U^T U$ then for any column 3-vector $v$ one has $v^T M v = v^T U^T U v=\|U v\|^2\geq 0$. Hence $M$ will be positive semidefinite by definition ($v^T M v\geq 0$ for all $v$.)

But an equivalent condition for a matrix to be PSD is that all the principal minors be nonnegative. These minors here are the diagonal elements, the 2-by-2 principal minors e.g. $\begin{vmatrix} 1 & x \\ x & 1\end{vmatrix}=1-x^2$, and the determinant. The first are 1's and therefore positive, the second is nonnegative since they're cosines of angles, and all that remains is $\det M = 1+2xyz-x^2-y^2-z^2$. So $M$ is PSD iff $\det M\geq 0$ with $(x,y,z)\in[-1,1]^3$.

Writing $f(x,y,z)=\det M$, you can check that $f(0,0,0)=1$, $f(1,1,1)=0$, and $f(1,1,-1)=-4$. So (1,1,-1) is not in the set of allowed cosines, (1,1,1) is on the boundary of said set, and (0,0,0) is inside the set.

As a special case, when $z=1$ one has $f(x,y,1)=2xy-x^2-y^2=-(x-y)^2$ which is nonnegative when $x=y$ i.e. the line segment $(x,y,z)=(t,t,1)$ with $t\in[-1,1]$ satisfies $f=0$.

That connects the corners $(1,1,1)$ and $(-1,-1,1)$. There will be five other such line segments which run between the four allowed vertices.

And by convexity of PSD matrices one concludes that the tetrahedron generated by the four allowed vertices is definitely in the set. So a natural question is whether there's any points outside of it.

and the answer turns out to be yes: for instance, the point $(-1/2,-1/2,-1/2)$ will be outside.

Secret

15:09

One thing I am currently thinking about is whether I can use $M$ to map some vector $v$ in $\Bbb{R}^3$ and demand that the image must be within the cube. But I need to think about what actually will the map $M : \Bbb{R}^3 \to [-1,1]^3$ mean, bcause if this can be made sense, I can get a matrix inequality thus allowing me to row reduce it to find the constraints on my input vectors $v$

Semiclassical

Right. And this has a geometric meaning as well

Let me consider spherical coordinates on the unit sphere except with $e_x=[1,0,0]^T$ as the axis rather than $e_z$

Akiva Weinberger

I don't know what this is but it looks cool

Source - https://en.wikipedia.org/wiki/Sweep_line_algorithm

Semiclassical

something something voronoi?

Akiva Weinberger

Yeah, but I don't know how the algorithm works

Semiclassical

"Since then, this approach has been used to design efficient algorithms for a number of problems, such as construction of the Voronoi diagram (Fortune's algorithm) and the Delaunay triangulation or Boolean operations on polygons."

cool

Then I want $x=\hat{a}\cdot \hat{b}=b_x$...ugh, I'm regretting calling the dot product as $x$ now

But anyways. In terms of the sphere, this is just a line of latitude. So we want to pick $\hat{b}$ to be some point on a prescribed line of latitude. Similarly we want $y=\hat{a}\cdot \hat{c}=c_x$

we might as well pick the azimuthal angle of the first one to be 0

for the second, though, we need to pick it in such a way that $z=\hat{b}\cdot\hat{c}$

Akiva Weinberger

15:26

Two ears theorem

In geometry, the two ears theorem states that every simple polygon with more than three vertices has at least two ears, vertices that can be removed from the polygon without introducing any crossings. The two ears theorem is equivalent to the existence of polygon triangulations. It is frequently attributed to Gary H. Meisters, but was proved earlier by Max Dehn. == Statement of the theorem == An ear of a polygon is defined as a vertex v such that the line segment between the two neighbors of v lies entirely in the interior of the polygon. The two ears theorem states that every simple polygon has...

Cool

0celo7

@Semiclassical ugh people brought up valid criticisms of my proof yesterday

Semiclassical

and that choice of azimuthal angle $\phi$ will correspond to the angle at vertex $\hat{a}=e_x$ of the spherical triangle

0celo7

but I can't figure out exactly why they're wrong

I know they're wrong, but the fact that they are is strange

Semiclassical

They can't be right, but they seem like they should be?

That's always frustrating.

0celo7

Mhm

Semiclassical

15:29

I meant to ask. How much are you actually planning to do with the rearrangements business?

Was it something you were actually going to explore or just a brief diversion?

0celo7

Brief diversion for an Appendix in my thesis and a couple seminar talks

Akiva Weinberger

Why would the dual of a triangulation be a tree?

0celo7

I would love to explore it given more time

In fact that old book by Polya and Szego looks fascinating

My advisor thinks I should give a talk on geometric measure theory in the undergrad seminar. So I might be learning more about this

Akiva Weinberger

Anthropomorphic polygon

In geometry, an anthropomorphic polygon is a simple polygon with precisely two ears and one mouth. That is, for exactly three polygon vertices, the line segment connecting the two neighbors of the vertex does not cross the polygon. For two of these vertices (the ears) the line segment connecting the neighbors forms a diagonal of the polygon, contained within the polygon. For the third vertex (the mouth) the line segment connecting the neighbors lies outside the polygon, forming the entrance to a concavity of the polygon. Every simple polygon has at least two ears (this is the two ears theorem)...

Lol

Secret

@AkivaWeinberger I have the following ad hoc "partial step": Suppose the lattice of polygons $S$ has an invariant $I$ that is translation invariant and additive. Then make a copy of $S$ by translating $S$ up by one unit. Now since $I$ is invariant and additive, you now have $2I$. We can then use an induction argument to conclude that in general the invariant is $nI$ for all integers $n$. I am still thinking how it relates to the area map

Semiclassical

15:40

@0celo7 cool

one thing i realized while doing the MMA is that while you can't easily visualize u(x) if you're in higher dimensions, you can visualize u(|x|) if it's spherically symmetric (but not necessarily decreasing)

so probably that's by necessity a good special case

Orpheus

Let $p,q$ be probability distributions on a finite set. Prove that

$\sum \frac{(p(x)-q(x))^2}{p(x)+q(x)} \leq 2\sum [p(x) log(\frac{p(x)}{q(x)})]$

Semiclassical

that looks weirdly asymmetric

i mean, the left side is unchanged if you exchange the two pmfs

but the right side isn't

@0celo7 though "radial u is good because I can visualize it regardless of dimensionality" is perhaps like the old story about a drunk looking under a streetlight for the keys he lost in the bar because that's where the light is good

0celo7

lol

it's because you decrease kinetic energy

$\|Du^\star\|_{L^p}\le \|Du\|_{L^p}$

for $p=2$ this says something about the expectation value of the Laplacian

Semiclassical

hmm

Not sure I know what kinetic energy corresponds to in a capacitance problem

0celo7

Oh I' not sure what the story is there

but people like to call $\|Du\|_{L^2}$ the kinetic energy

Semiclassical

15:55

hmm

i'm also not sure what $u$ is supposed to be here. It's scalar, so...electric potential?

Secret

The following quadratic form will allow me to do the direct mapping of $[a,b,c] \to [-1,1]^3$. However I need to think about how to set up the inequality:

Semiclassical

I tend to think of the mapping as $[a,b,c]\mapsto [a,b,c]^T [a,b,c]$ and then project to the three off-diagonal components.

Secret

$$\begin{pmatrix}a & b & c\end{pmatrix}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}\begin{pmatrix}a \\ b \\ c\end{pmatrix} = \begin{pmatrix}m \\ n \\ p\end{pmatrix}$$

Semiclassical

That can't be right. (a,b,c) is a row vector with column vector entries i.e. it's just a 3-by-3 matrix

Secret

Do I get the components to dot together when I do a "dot product" like this?
$\begin{pmatrix}a \\ b \\ c \end{pmatrix}\cdot \begin{pmatrix} b \\ c \\ a\end{pmatrix}$

because that's what I am trying to say in the above expression

wilkersmon

16:21

Can someone help me prove this?

I'm only allowed to use these properties:

I suspect I need to use contractions here

however i'm not entirely sure how they work

and how they would apply here

Secret

Ok, I think if I am not sloppy it should actually look like this
$\begin{pmatrix}\textbf{a} \\ \textbf{b} \\ \textbf{c}\end{pmatrix}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}\begin{pmatrix}\textbf{a} \\ \textbf{b} \\ \textbf{c}\end{pmatrix} = \begin{pmatrix}\textbf{m} \\ \textbf{n} \\ \textbf{p}\end{pmatrix}$
and then contract the resulting matrix in the direction of the rows

wilkersmon

@Secret what you wrote doesn't make sense

it's invalid matrix multiplication

Secret

so that $\textbf{a} \cdot \textbf{b} = \sum_{i=1}^3 m_i$ and similarly for others

typo

wilkersmon

what are you trying to achieve here?

Secret

$$\begin{pmatrix}\textbf{a}^T \\ \textbf{b}^T \\ \textbf{c}^T\end{pmatrix}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}\begin{pmatrix}\textbf{a}^T \\ \textbf{b}^T \\ \textbf{c}^T\end{pmatrix} = \begin{pmatrix}\textbf{m}^T \\ \textbf{n}^T \\ \textbf{p}^T\end{pmatrix}$$

tmr, will try to figure out how to rule out (1,1,-1) from this

Semiclassical

16:50

@Secret tbh I don't think that helps you much, since you still only want a few of those elements (the diagonal ones in this case I guess)

My personal suggestion is to write out what $\det([a,b,c]^T [a,b,c])$ would have to be in the (1,1,-1) case

(It's important that $[a,b,c]$ is a square matrix; if not, then things are a lot hairier as I can personally attest.)

On that note, I really wish the formula for $\det(A^T B)$ with $A,B$ of the same shape wasn't so bad when $A,B$ aren't square :(

feynhat

17:07

@TobiasKildetoft Metric spaces are Hausdorff right?

Daminark

@feynhat yeah, the idea is that you choose an open ball whose radius is at most half the distance from x and y around each

feynhat

@Daminark Thanks.

So, if $X$ is a metric space and $A$ is a subset of $X$ then, will there always be a sequence in $A$, converging to a limit point of $A$ in $X$.

wilkersmon

@feynhat Yes

Just form repeatedly smaller neighborhoods around that point

feynhat

Phew. Thanks.

wilkersmon

the condition of limit point guarantees that there are points in A in that neighborhood

feynhat

17:12

@wilkersmon Thanks.

That solves my problem.

I was actually trying to show that if two functions agree on a set, then they must also agree on its closure.

*continuous functions

wilkersmon

Right, and this would prove it

because a continuous function at a point is completely determined by a sequence converging to that point

Also, I just became curious

if we take $v w^{T}$, that's just like the tensor product of these two vectors, right?

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